Question

Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute its density.

Answer 1

The density of the molybdenum BCC crystal structure is 10.22 g/cm^3

Step-by-step explanation:

Density (D) of BCC crystal structure is given as:

D = (n × AW) ÷ (12.32r^3 × Na)

n is the number of atoms of molybdenum per unit cell = 2

AW is the atomic weight of molybdenum = 95.94 g/mol

r is the atomic radius of molybdenum = 0.1363 nm = 0.1363×10^-9 m = 0.1363×10^-9 × 100 = 1.363×10^-8 cm

Na is Avogadro's number = 6.022×10^23 atoms/mol

D = (2×95.94) ÷ [12.32 × (1.363×10^-8)^3 × 6.022×10^23] = 191.88 ÷ 18.78 = 10.22 g/cm^3

Answer 2

The density of the crystal is 10.22 g/cm³

Step-by-step explanation:

Step 1: Data given

Molybdenum (Mo) has a BCC crystal structure

atomic radius = 0.1363 nm

atomic weight of 95.94 g/mol

Step 2: Calculate density of a bcc crystal

Density ρ= (nA *Mo)/(Vc*Na)

⇒For BCC, n = 2 atoms/unit cell, and realizing that V c = a ³

a = 4R/√3

Vc =(4R/√3)³

AMo = atomic weight = 95.94 g/mol

ρ = (n*A Mo)/((4R/√3)³ *Na)

⇒with n = 2 atoms/unit cell

⇒ with AMo = 95.94 g/mol

⇒ with Vc =(4R/√3)³ = (4*0.1363*10-7 cm)³ /(√3)³

⇒with Na = 6.022*10^23 atoms/mol

ρ =10.22 g/cm³

The density of the crystal is 10.22 g/cm³